Question

The lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ are parallel. If $\angle BAE=54^{\circ}$ and $\angle DCE=25^{\circ}$, what is the measure of $\angle AEC$?
Use an asymptote convertor if needed:

[asy]
size( 200 ) ;

pair A = (0,0) ;
pair B = (0,2) ;
pair C = (3,0.5) ;
pair D = (3,2.5) ;
pair ptE = extension( A , A+dir(90-54) , C , C+dir(90+25) ) ;

draw( (0,-0.5)--(0,3) , linewidth(1.3) ) ;
draw( (3,-0.5)--(3,3) , linewidth(1.3) ) ;
draw( A--ptE--C ) ;
dot(Label( "$A$" , A , W )) ;
dot(Label( "$B$" , B , W )) ;
dot(Label( "$C$" , C , E )) ;
dot(Label( "$D$" , D , E )) ;
dot(Label( "$E$" , ptE , N )) ;

label( "$54^{\circ}$" , A , 5dir(70) ) ;
label( "$25^{\circ}$" , C , 8dir(100) ) ;
[/asy]

Answer 1

Answer:

  79°

Step-by-step explanation:

Label the intersection of AE and CD point F. Then angle CFE is an alternate interior angle with angle BAE, so is congruent. The angle of interest is an exterior angle to triangle CFE, so is the sum of remote interior angles CFE (54°) and DCE (25°). That sum is ...

  m∠AEC = 54° +25° = 79°